A simple generalization would be to replace geometric brownian motion with geometric fractional brownian motion w/ H in (0,1) ==> GBM is just a GFBM w/ H=1/2. From what I can recall, the only difference in the log- (fractional) normal distribution is sigma==> sigma*(T-t)^H-0.5.
@adang23 yes, thank you, of course I agree that is more correct! (i don't think it impacts anything subsequently ... I confuse myself sometimes because "expected return" mu can be defined either as mu-variance/2 or just mu but clearly your are right about the LN() distribution).
@akathetruthteller yes E[(e^X)] = e^[mu + sigma^2/2] but S(t) = S(0)e^[(mu - sigma^2/2)T+sigma*sqrt(T)*z] reduces to E[St] = S(0)*e^(mu*T) for the mean which is greater than the median. Not so much as shift as, to paraphrase Culp, how you define the drift
I like the video but I might have to point out something here. The result and definition of Mean are not accurate here. ln(S_T/S_0)~N(u, sigma^2), then when you calculate expectation(mean) of S_T, it should be integral of (g(x)f(x)dx), which equals to S_0*e^(u-(sigma^2)/2) i.e. the result of the median in the video. Then you use the expectation to calculate the confidence interval and VaR with the approximation to normal distribution. The definition of VaR and confidence interval both deal with the expectation(mean) instead of the median.
man, you should double check. i think for lognormal RV. E(X) =exp( mu + sigma^2/2) and median = exp(mu). you are implying a shift of calculation somehow.
A simple generalization would be to replace geometric brownian motion with geometric fractional brownian motion w/ H in (0,1) ==> GBM is just a GFBM w/ H=1/2. From what I can recall, the only difference in the log- (fractional) normal distribution is sigma==> sigma*(T-t)^H-0.5.
@adang23 yes, thank you, of course I agree that is more correct! (i don't think it impacts anything subsequently ... I confuse myself sometimes because "expected return" mu can be defined either as mu-variance/2 or just mu but clearly your are right about the LN() distribution).
Nice example. May be nice if you showed everyone how to actually calculate drift (return), as well, rather than just assume, to use when appropriate.
@akathetruthteller yes E[(e^X)] = e^[mu + sigma^2/2] but S(t) = S(0)e^[(mu - sigma^2/2)T+sigma*sqrt(T)*z] reduces to E[St] = S(0)*e^(mu*T) for the mean which is greater than the median. Not so much as shift as, to paraphrase Culp, how you define the drift
I think it is more accurate to say ln(S_T/S_0) follows a N(mu-0.5sigma^2,sigma^2) distribution, rather than N(mu, sigma^2) distribution...
Hey, thank you so much for the great video. Can you give me a hint in which bibliography sources I can get these formulas? Thanks a bunch :D
I like the video but I might have to point out something here. The result and definition of Mean are not accurate here. ln(S_T/S_0)~N(u, sigma^2), then when you calculate expectation(mean) of S_T, it should be integral of (g(x)f(x)dx), which equals to S_0*e^(u-(sigma^2)/2) i.e. the result of the median in the video. Then you use the expectation to calculate the confidence interval and VaR with the approximation to normal distribution. The definition of VaR and confidence interval both deal with the expectation(mean) instead of the median.
man, you should double check. i think for lognormal RV. E(X) =exp( mu + sigma^2/2)
and median = exp(mu). you are implying a shift of calculation somehow.